Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{4r^2 + 12r}{-r^2 + 2r + 15} \times \dfrac{r^2 - 8r + 16}{2r^2 - 8r} $
Explanation: First factor out any common factors. $q = \dfrac{4r(r + 3)}{-(r^2 - 2r - 15)} \times \dfrac{r^2 - 8r + 16}{2r(r - 4)} $ Then factor the quadratic expressions. $q = \dfrac {4r(r + 3)} {-(r + 3)(r - 5)} \times \dfrac {(r - 4)(r - 4)} {2r(r - 4)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {4r(r + 3) \times (r - 4)(r - 4) } { -(r + 3)(r - 5) \times 2r(r - 4)} $ $q = \dfrac {4r(r - 4)(r - 4)(r + 3)} {-2r(r + 3)(r - 5)(r - 4)} $ Notice that $(r + 3)$ and $(r - 4)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {4r(r - 4)(r - 4)\cancel{(r + 3)}} {-2r\cancel{(r + 3)}(r - 5)(r - 4)} $ We are dividing by $r + 3$ , so $r + 3 \neq 0$ Therefore, $r \neq -3$ $q = \dfrac {4r\cancel{(r - 4)}(r - 4)\cancel{(r + 3)}} {-2r\cancel{(r + 3)}(r - 5)\cancel{(r - 4)}} $ We are dividing by $r - 4$ , so $r - 4 \neq 0$ Therefore, $r \neq 4$ $q = \dfrac {4r(r - 4)} {-2r(r - 5)} $ $ q = \dfrac{-2(r - 4)}{r - 5}; r \neq -3; r \neq 4 $